CSCE 411 Design and Analysis of Algorithms Andreas Klappenecker TexPoint fonts used in EMF. Read the TexPoint manual before you delete this box.: AAAA

Goal of this Lecture Recall the basic asymptotic notations such as Big Oh, Big Omega, Big Theta, and little oh. Recall some basic properties of these notations Give some motivation why these notions are defined in the way they are. Give some examples of their usage.

Summary Let g: N->C be a real or complex valued function on the natural numbers. O(g) = { f: N-> C | u>0 n0 N |f(n)| <= u|g(n)| for all n>= n0 } (g) = { f: N-> C | d>0 n0 N d|g(n)| <= |f(n)| for all n>= n0 } (g) = { f: N-> C | u,d>0 n0 N d|g(n)| <= |f(n)| <= u|g(n)| for all n>= n0 } o(g) = { f: N-> C | limn-> |f(n)|/|g(n)| = 0 }

Time Complexity Estimating the time-complexity of algorithms, we simply want count the number of operations. We want to be independent of the compiler used, ignorant about details about the number of instructions generated per high-level instruction, independent of optimization settings, and architectural details. This means that performance should only be compared up to multiplication by a constant. We want to ignore details such as initial filling the pipeline. Therefore, we need to ignore the irregular behavior for small n.

Big Oh

Big Oh Notation Let f,g: N -> R be function from the natural numbers to the set of real numbers. We write f  O(g) if and only if there exists some real number n0 and a positive real constant u such that |f(n)| <= u|g(n)| for all n in S satisfying n>= n0

Big Oh Let g: N-> C be a function. Then O(g) is the set of functions O(g) = { f: N-> C | there exists a constant u and a natural number n0 such that |f(n)| <= u|g(n)| for all n>= n0 }

Notation We have O(n2)  O(n3) but it usually written as O(n2) = O(n3) This does not mean that the sets are equal!!!! The equality sign should be read as ‘is a subset of’.

Notation We write n2 = O(n3), [ read as: n2 is contained in O(n3) ] But we never write O(n3) = n2

Example O(n2)

Big Oh Notation The Big Oh notation was introduced by the number theorist Paul Bachman in 1894. It perfectly matches our requirements on measuring time complexity. Example: 4n3+3n2+6 in O(n3) The biggest advantage of the notation is that complicated expressions can be dramatically simplified.

Quiz Does O(1) contain only the constant functions?

Big versus Little Oh O(g) = { f: N-> C | u>0 n0 N |f(n)| <= u|g(n)| for all n>= n0 } o(g) = { f: N-> C | limn-> |f(n)|/|g(n)| = 0 }

Limit Let (xn) be a sequence of real numbers. We say that  is the limit of this sequence of numbers and write  = limn-> xn if and only if for each  > 0 there exists a natural number n0 such that |xn - |<  for all n >= n0

? !

Limit – Again! Let (xn) be a sequence of real numbers. We say that  is the limit of this sequence of numbers and write  = limn-> xn if and only if for each  > 0 there exists a natural number n0 such that |xn - |<  for all n >= n0

How do we prove that g = O(f)?

Quiz It follows that o(f) is a subset of O(f). Why?

Quiz What does f = o(1) mean? Hint: o(g) = { f: N-> C | limn-> |f(n)|/|g(n)| = 0 }

Quiz Some computer scientists consider little oh notations too sloppy. For example, 1/n+1/n2 is o(1) but they might prefer 1/n+1/n2 = O(1/n). Why is that?

Limits? There are no Limits! The limit of a sequence might not exist. For example, if f(n) = 1+(-1)n then limn-> f(n) does not exist.

Least Upper Bound (Supremum) The supremum b of a set of real numbers S is the defined as the smallest real number b such that b>=x for all s in S. We write s = sup S. sup {1,2,3} = 3, sup {x : x2 <2} = sqrt(2), sup {(-1)^n – 1/n : n>=0 } = 1.

The Limit Superior The limit superior of a sequence (xn) of real numbers is defined as lim supn -> xn = limn -> ( sup { xm : m>=n}) [Note that the limit superior always exists in the extended real line (which includes ), as sup { xm : m>=n}) is a monotonically decreasing function of n and is bounded below by any element of the sequence.]

The Limit Superior The limit superior of a sequence of real numbers is equal to the greatest accumulation point of the sequence.

Necessary and Sufficient Condition

Big Omega

Big Omega Notation Let f, g: N-> R be functions from the set of natural numbers to the set of real numbers. We write g  (f) if and only if there exists some real number n0 and a positive real constant C such that |g(n)| >= C|f(n)| for all n in N satisfying n>= n0.

Big Omega Theorem: f(g) iff lim infn->|f(n)/g(n)|>0. Proof: If lim inf |f(n)/g(n)|= C>0, then we have for each >0 at most finitely many positive integers satisfying |f(n)/g(n)|< C-. Thus, there exists an n0 such that |f(n)|  (C-)|g(n)| holds for all n  n0, proving that f(g). The converse follows from the definitions.

Big Theta

Big Theta Notation Let S be a subset of the real numbers (for instance, we can choose S to be the set of natural numbers). If f and g are functions from S to the real numbers, then we write g  (f) if and only if there exists some real number n0 and positive real constants C and C’ such that C|f(n)|<= |g(n)| <= C’|f(n)| for all n in S satisfying n>= n0 . Thus, (f) = O(f)  (f)

Examples

Sums 1+2+3+…+n = nn+1/2 12 +22 +32 +…+n2 = n(n+1)(2n+1)/6 We might prefer some simpler formula, especially when looking at sum of cubes, etc. The first sum is approximately equal to n2/2, as n/2 is much smaller compared to n2/2 for large n. The first sum is approximately equal to n3/3 plus smaller terms.

Approximate Formulas ( complicated function of n) = (simple function of n) + (bound for the size of the error in terms of n)

Approximate Formulas Instead of 12 +22 +32 +…+n2 = n3/3+ n2/2 + n/6 we might write 12 +22 +32 +…+n2 = n3/3 + O(n2)

Approximate Formulas If we write f(n) = g(n)+O(h(n)), then this means that there exists a constant u>0 and a natural number n0 such that |f(n)-g(n)| <= u|h(n)| for all n>=n0.

Bold Conjecture 1k +2k +3k +…+nk = nk+1/(k+1) + O(nk)

Proof Write S(n) = 1k +2k +3k +…+nk We try to estimate S(n). S(n) <1n+1xk dx nk S(n) <(n+1)k/(k+1) 3k 2k 1k

Proof Write S(n) = 1k +2k +3k +…+nk We try to estimate S(n). S(n) >0n xk dx S(n) > nk+1/(k+1) 3k 2k 1k

Proof We have shown that nk+1/(k+1) < 1k +2k +3k +…+nk < (n+1)k+1/(k+1). Let’s subtract nk+1/(k+1) to get 0< 1k +2k +3k +…+nk – nk+1/(k+1) < ((n+1) k+1-nk+1 )/(k+1)

Proof <= (some mess involving k) nk It follows that ((n+1) k+1-nk+1 )/(k+1) =(k+1)-1 i=0k+1C(k+1,i)ni- nk+1 <= i=0k C(k+1,i)nk // simplify and enlarge terms <= (some mess involving k) nk It follows that S(n) - nk+1/(k+1) < (mess involving k) nk

End of Proof Since the mess involving k does not depend on n, we have proven that 1k +2k +3k +…+nk = nk+1/(k+1) + O(nk) holds!

Harmonic Number The Harmonic number Hn is defined as Hn = 1+1/2+1/3+…+1/n. We have Hn = ln n +  + O(1/n) where  is the Euler-Mascheroni constant =0.577…

log n! Recall that 1! = 1 and n! = (n-1)! n. Theorem: log n! = (n log n) Proof: log n! = log 1 + log 2 + … + log n <= log n + log n + … + log n = n log n Hence, log n! = O(n log n).

log n! On the other hand, log n! = log 1 + log 2 + … + log n >= log ((n+1)/2) + … + log n >= ((n+1)/2) log ((n+1)/2) >= n/2 log(n/2) = (n log n) For the last step, note that lim infn-> (n/2 log(n/2))/(n log n) = ½.

Reading Assignment Read Chapter 1-3 in [CLRS] Chapter 1 introduces the notion of an algorithm Chapter 2 analyzes some sorting algorithms Chapter 3 introduces Big Oh notation